In accordance with Theorem 2.5 [1]} , we get \( W ^{\ast }=T_{t^{\ast }}\)  (since \(W^{\ast }=\tilde{W}^{\ast }\) ).
Now if we denote \(t_{1}:=t^{\ast },\)  then it is easy to calculate \(\mathfrak {k} =-\mathfrak {k}_{1}.\)  Since \(t\)  is sectorial, than \(|\mathfrak {k}_{1}|\le \tan \theta \cdot \mathfrak {h}.\)  Hence, in accordance with Lemma 3.1 [1]}, we get \(\mathfrak {k} [u,v]=(BH^{1/2}u,H^{1/2}v),\,\mathfrak {k}_{1}[u,v]=-(BH^{1/2}u,H^{1/2}v),\,u,v\in \mathrm {D}(H^{1/2}),\)  where \(B\in \mathcal {B}(\mathfrak {H}) \)  is a symmetric operator. Let us prove that \(B\)  is selfadjoint. Note that in accordance with Lemma 3.1 [1]} \(\mathrm {D}(B)=\mathrm {R}(H^{1/2}),\)  in accordance with Theorem 2.1 [1]}, we have \((Hf,f)_{\mathfrak {H}}\ge C_{2}/C_{0}\Vert f\Vert _{\mathfrak {H}}^{2},\,f\in \mathrm {D}(H),\)  using the reasonings of Theorem 5 [5]}, we conclude that \(\mathrm {R}(H^{1/2})=\mathfrak {H}\)  i.e. \(\mathrm {D}(B)=\mathfrak {H}.\)  Hence \(B\)  is selfadjoint. Using Lemma 3.2 [1]}, we obtain a representation \(\tilde{W}=H^{1/2}(I+iB) H^{1/2},\,W^{\ast }=H^{1/2}(I-iB) H^{1/2}.\)  Noting the fact \(\mathrm {D}(B)=\mathfrak {H},\)  we can easily obtain \( (I\pm iB)^{\ast }= I \mp iB.\)  Since \(B\)  is selfadjoint, then
\(\mathrm {Re}([I\pm iB]f,f)_{\mathfrak {H}}=\Vert f\Vert ^{2}_{\mathfrak {H}}.\)  Using this fact and applying Theorem 3.2 [1]}, we conclude that \(\mathrm {R}(I\pm iB)\)  is a closed set. Since \(\mathrm {N}(I\pm iB)=0,\)  then \(\mathrm {R}(I \mp iB)=\mathfrak {H}\)  (see (3.2) [1]}). Thus, we obtain \((I\pm i B)^{-1}\in \mathcal {B}(\mathfrak {H}).\)  Taking into account the above facts, we get \(R_{\tilde{W}}=H^{-1/2}(I+iB)^{-1} H^{-1/2},\,R_{W^{\ast }}=H^{-1/2}(I-iB)^{-1} H^{-1/2}.\)  In accordance with the well-known theorem (see Theorem 5 [9]}), we have \(R^{\ast }_{\tilde{W} }=R_{W^{\ast }}.\)  Note that the relations \((I\pm i B)\in \mathcal {B}(\mathfrak {H}),\,(I\pm i B)^{-1}\in \mathcal {B}(\mathfrak {H}),\,H^{-1/2}\in \mathcal {B}(\mathfrak {H})\)  allow as to obtain the following formula by direct calculations
\(\mathfrak {Re} R _{\tilde{W} } =\frac{1}{2}H^{-1/2}(I+ B^{2})^{-1} H^{-1/2}.\) 
